# Magnifying power of a microscope for a farsighted person

I was doing some problems on optical instruments and in one of the questions it was asked that if a simple microscope(magnifying glass) has a magnifying power 5X for a normal relaxed eye.. what will be its magnifying power for a relaxed farsighted eye whose near point is 40cm? I have got the answer ( ... quite easy I know) and that comes out to be 8X . But does this mean that a farsighted person would see a larger image of an object as compared to a normal(healthy) relaxed eye?? I am not getting the intuition for this .... How this happens? Thank you...

It is a silly question. For magnifications of 5X one uses a magnifying glass or loupe, not a compound microscope.

But let's regard a magnifying glass with a focal length of 25 cm. According to standard rules, it just has magnifying power of unity.

But it is very useful for a farsighted person with a near point of for example 1 meter. It allows Sherlock Holmes to bring the item (newspaper, fingerprints) closer to his eyes, just as close as when he was young.

• So, are you saying a far sighted person (having LDDV*>25 cm) can see a bigger image than a normal person (having LDDV*=25 cm)? *LDDV-Least Distance of Distinct Vision – user14250 Jan 21 '20 at 5:33
• @GuruVishnu No, but a magnifying glass makes a larger difference for a far-sighted person. They cannot get a good image with the object closer to the eye. – user137289 Jan 21 '20 at 8:02
• Thank you for your reply. Then how could it be "very useful for a far sighted person"? I've learnt that the magnifying power for normal adjustment is given by $D/f$ and so it must increase with increase in $D$ i.e., least distance of distinct vision. And this is how I reasoned the statement in your answer. If angular magnification increases with $D$ then it seems a person with larger $D$ will see a bigger image contradictory to your previous comment. Am I missing anything important? – user14250 Jan 21 '20 at 8:11
• @GuruVishnu Consider a far-sighted person with LLDV of one meter. At that distance, newsprint may be too small to read. A magnifier with $f = 25$ cm allows that person to read their newspaper at 25 cm instead. So that would be four times as large as with the unaided eye, but the image on his retina has the same size as for a normal eye. – user137289 Jan 21 '20 at 8:57

The magnification of a compound microscope is the product of the objective magnification and the eyepiece magnification. Eyepiece magnification is defined as m= 250/f, where f is the focal length of the eyepiece in mm. For example, an eyepiece with a focal length of 25 mm will have a magnification of 10x. The value of 250 mm comes this way: it is chosen as the standard working distance for comparison of the unmagnified image to the magnified one. So, if you compare a target at 250 mm to one view with an focal length =50 mm lens, the one viewed with the lens will appear about 5X as big.

If, as I suspect, you said the magnification for the hyperope is 400/f, then one would get 400/50 or mag of 8. To my knowledge, no one does this calculation this way. The reason is it would cause confusion: the compound magnification of a system should not depend on who is going to use it.

As to the apparent size of the image, hyperopes usually have a shorter eyeball length. I don't think that you can really compare the apparent angular size for a myope and a hyperope - absolute calibration is pointless, since these people can't switch retinas or eye lenses. (This is unlike cameras, where you know that the image sensor is say, 10 mm x 10 mm, and then one can compare two lenses with different focal lengths. The different focal lengths will produce images of different sizes. With cameras, people do pair different focal length lenses with different size sensors, and you need to know ahead of time what will happen.)

• I think the question "does this mean that a farsighted person would see a larger image of an object as compared to a normal eye?" is best answered by comparing the angular image size, instead of the linear size of the image on the retina. Then the answer will be that the angular image size is the same for both persons. In practice, a person can easily determine the angular image size, by alternatingly looking at the image and a measuring rod a few meters away. The angular size can be calculated from the apparent size at the rod. – jkien Dec 9 '18 at 12:35